Question

What is the expression for the capacitance of a slab of material having dielectric constant K which have same area as that of plates of parallel plate capacitor but has thickness of $\dfrac{d}{2}$ where, $d$ is distance between the plates when slab is inserted between plates of capacitor.

Answer 1

Hint There is vacuum between two plates so, use this expression to find out the capacitance of capacitor
$C = \dfrac{{{\varepsilon _0}A}}{d}$ (where, A is the area of plates)
If the dielectric is inserted, so, the electric field in the dielectric is
 $E = \dfrac{{{E_0}}}{K}$

Complete step-by-step answer:
Capacitor is the device which holds the charge. Capacitance is the capacity of a capacitor to hold the charge. Its S.I unit is Farad. It also opposes the variation of voltage.
When there is vacuum between two plates, then the expression used is
$C = \dfrac{{{\varepsilon _0}A}}{d}$
If a capacitor is connected to the battery, the electric field produced is ${E_0}$.
Now, if we insert a slab of thickness $\dfrac{d}{2}$, the electric field reduces to $E$. The separation between two plates is divided in two parts, for distance there is an electric field $E$and for distance remaining $\left( {d - t} \right)$ the electric field is ${E_0}$.
Let $V$be the potential difference between two plates. Therefore,
$
  V = Et + {E_0}\left( {d - t} \right) \\
  V = E\dfrac{d}{2} + {E_0}\dfrac{d}{2} \\
 $
Substituting the value of $t = \dfrac{d}{2}$ in the above equation
$V = \dfrac{d}{2}\left( {E + {E_0}} \right)$
$
  V = \dfrac{d}{2}\left( {\dfrac{{{E_0}}}{K} + {E_0}} \right) \\
  V = \dfrac{{d{E_0}}}{{2K}}(K + 1) \cdots (i) \\
 $
Because the value of $\dfrac{{{E_0}}}{E} = K$
The value of
$
  {E_0} = \dfrac{\sigma }{{{\varepsilon _0}}} \\
  \because \sigma = \dfrac{q}{A} \\
  \therefore {E_0} = \dfrac{q}{{{\varepsilon _0}A}} \\
 $
Putting this value of ${E_0}$in equation $(i)$we get,
$ \Rightarrow V = \dfrac{d}{{2K}}\dfrac{q}{{{\varepsilon _0}A}}(K + 1) \cdots (ii)$
We know that,
Capacitance is equal to $\dfrac{q}{v}$, $C = \dfrac{q}{v} \cdots (iii)$
Putting the value of equation $(ii)$in $(iii)$, we get
$C = \dfrac{{2K{\varepsilon _0}A}}{{(K + 1)d}}$

Note As we know that, $V\propto R$
$
  V = \dfrac{1}{C}R \\
  Q = CV \\
   \Rightarrow C = \dfrac{Q}{V} \\
 $(where, $C$ is the Capacitance)
$V = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{Q}{R}$
$
  \therefore C = \dfrac{Q}{V} = \dfrac{Q}{{\dfrac{1}{{4\pi {\varepsilon _0}\dfrac{Q}{R}}}}} \\
  C = 4\pi {\varepsilon _0}R \\
 $
Capacitance of earth is $711\mu F$(micro-Farad).